Hypothesis tests

Appendices

Hypothesis Test: Difference Between Paired Means

This lesson explains how to conduct a hypothesis test for the
difference between
paired means. The test procedure, called the
matched-pairs t-test,
is appropriate when the following conditions are met:

The population data are slightly
skewed,
unimodal, without outliers,
and the sample size is 16 to 40.

The sample size is greater than 40, without outliers.

This approach consists of four steps: (1) state the hypotheses,
(2) formulate an analysis plan, (3) analyze sample data, and
(4) interpret results.

State the Hypotheses

Every hypothesis test requires the analyst
to state a
null hypothesis
and an
alternative hypothesis. The hypotheses are stated in such
a way that they are mutually exclusive. That is, if one is
true, the other must be false; and vice versa.

The hypotheses concern a new variable d, which is based on the
difference between paired values from two data sets.

d = x1 - x2

where x1 is the value of variable x in the first data set,
and x2 is the value of the variable from the second data set
that is paired with x1.

The table below shows three sets of null and alternative hypotheses.
Each makes a statement about how the true difference in
population values μd is related to some hypothesized value D.
(In the table, the symbol ≠ means " not equal to ".)

Set

Null hypothesis

Alternative hypothesis

Number of tails

1

μd= D

μd ≠ D

2

2

μd> D

μd < D

1

3

μd< D

μd > D

1

The first set of hypotheses (Set 1) is an example of a
two-tailed test, since an extreme value on either side of the
sampling distribution would cause a researcher to reject the null
hypothesis. The other two sets of hypotheses (Sets 2 and 3) are
one-tailed tests, since an extreme value on only one side of the
sampling distribution would cause a researcher to reject the
null hypothesis.

Formulate an Analysis Plan

The analysis plan describes
how to use sample data to accept or reject the null
hypothesis. It should specify the following elements.

Significance level. Often, researchers choose
significance levels
equal to
0.01, 0.05, or 0.10; but any value between 0 and
1 can be used.

Test method. Use the
matched-pairs t-test
to determine whether the difference between sample means
for paired data is
significantly different from the hypothesized difference
between population means.

Analyze Sample Data

Using sample data, find the standard deviation, standard error,
degrees of freedom,
test statistic, and the P-value associated with the test statistic.

where di is the difference for pair i,
d is the sample mean of the differences,
and n is the number of paired values.

Standard error. Compute the
standard error (SE)
of the sampling distribution of d.

SE =
sd *
sqrt{ ( 1/n ) * [ (N - n) / ( N - 1 ) ] }

where sd is the standard deviation of the sample
difference, N is the number of matched pairs in the population, and
n is the number of matched pairs in the sample. When the population size is much
larger (at least 20 times larger) than the sample size, the
standard error can be approximated by:

Test statistic. The test statistic is a t statistic (t) defined by
the following equation.

t = [ (x1
- x2) - D ]
/ SE
= (d - D) / SE

where
x1 is the mean of sample 1,
x2 is the mean of sample 2,
d is the mean difference between
paired values in the sample,
D is the hypothesized difference between population means,
and SE is the standard error.

P-value. The P-value is the probability of observing a
sample statistic as extreme as the test statistic. Since the
test statistic is a t statistic, use the
t Distribution Calculator
to assess the probability associated with the t statistic, having
the degrees of freedom computed above. (See the
sample problem at the end of this lesson for guidance on how this
is done.)

Interpret Results

If the sample findings are unlikely, given
the null hypothesis, the researcher rejects the null hypothesis.
Typically, this involves comparing the P-value to the
significance level,
and rejecting the null hypothesis when the P-value is less than
the significance level.

Test Your Understanding

Problem

Forty-four sixth graders were randomly selected from a school district.
Then, they were divided into 22 matched pairs, each
pair having equal IQ's. One member of each pair was randomly
selected to receive special training. Then,
all of the students were given an IQ test. Test results are
summarized below.

Pair

Training

No training

Diff, d

(d - d)2

1

95

90

5

16

2

89

85

4

9

3

76

73

3

4

4

92

90

2

1

5

91

90

1

0

6

53

53

0

1

7

67

68

-1

4

8

88

90

-2

9

9

75

78

-3

16

10

85

89

-4

25

11

90

95

-5

36

Pair

Training

No training

Diff, d

(d - d)2

12

85

83

2

1

13

87

83

4

9

14

85

83

2

1

15

85

82

3

4

16

68

65

3

4

17

81

79

2

1

18

84

83

1

0

19

71

60

11

100

20

46

47

-1

4

21

75

77

-2

9

22

80

83

-3

16

Σ(d - d)2 = 270
d = 1

Do these results provide evidence that the special training
helped or hurt student performance? Use an 0.05 level of
significance. Assume that the mean differences are approximately
normally distributed.

Solution

The solution to this problem takes four steps:
(1) state the hypotheses, (2) formulate an analysis plan,
(3) analyze sample data, and (4) interpret results.
We work through those steps below:

State the hypotheses. The first step is to
state the null hypothesis and an alternative hypothesis.

Null hypothesis: μd = 0

Alternative hypothesis:
μd ≠ 0

Note that these hypotheses constitute a two-tailed test.
The null hypothesis will be rejected if the
difference between sample means
is too big or if it is too small.

Formulate an analysis plan. For this analysis,
the significance level is 0.05. Using sample data, we will
conduct a
matched-pairs t-test
of the null hypothesis.

Analyze sample data. Using sample data, we
compute the standard deviation of the differences (s),
the standard error (SE) of the mean difference,
the degrees of freedom (DF),
and the t statistic test statistic (t).

s = sqrt [
(Σ(di - d)2
/ (n - 1) ]

s = sqrt[ 270/(22-1) ] = sqrt(12.857) = 3.586

SE = s / sqrt(n) = 3.586 / [ sqrt(22) ]

SE = 3.586/4.69 = 0.765

DF = n - 1 = 22 -1 = 21

t = [ (x1
- x2) - D ]
/ SE

t
= (d - D)/ SE
= (1 - 0)/0.765 = 1.307

where
di is the observed difference for pair i,
d is mean difference between
sample pairs,
D is the hypothesized mean difference between population pairs,
and n is the number of pairs.

Since we have a
two-tailed test, the P-value is the probability that a
t statistic having 21 degrees of freedom is more extreme than 1.307;
that is, less than -1.307 or greater than 1.307.

Interpret results. Since the P-value (0.206) is
greater than the significance level (0.05), we cannot reject the
null hypothesis.

Note: If you use this approach on an exam, you may also want to mention
why this approach is appropriate. Specifically, the approach is
appropriate because the sampling method was simple random sampling,
the samples consisted of paired data, and the mean differences were
normally distributed. In addition, we used the approximation formula
to compute the standard error, since the sample size was small
relative to the population size.